The concept of projectile motion is a cornerstone of physics, illustrating how objects move under gravity’s influence. This often involves solving problems related to vertical and horizontal displacement, and it becomes particularly interesting when dealing with multiple objects. Today, we’ll examine how to determine when two stones launched at different times and velocities meet, specifically when their vertical positions, or match vertical heights, coincide.

---

---

Projectile Motion — When Two Stones Reach the Same Height

Projectile motion often requires analyzing simultaneous vertical positions of two objects. In this case, we explored when two stones, thrown at different times and velocities, align at the same vertical height under gravity. This problem showcases how kinematics and algebra blend to solve real-world scenarios.

Equations of Motion

• Stone A (launched at ##t=0##): ### y_A(t) = 18t – 4.9t^2 ###
• Stone B (launched at ##t=2.5## s): ### y_B(t) = 32(t-2.5) – 4.9(t-2.5)^2 ###

Solving for Equal Heights

Set the equations equal: ### 18t – 4.9t^2 = 32(t-2.5) – 4.9(t-2.5)^2. ###

Expanding and simplifying yields: ### 0 = 38.5t – 110.625. ###

Thus: ### t = \tfrac{110.625}{38.5} \approx 2.87\;\text{s}. ###

Common Height

Substitute ##t=2.87## into Stone A’s equation: ### y_A(2.87) = 18(2.87) – 4.9(2.87)^2 \approx 11.4\;\text{m}. ###

This matches Stone B’s calculation, confirming accuracy.

Results

• Meeting time: ##t \approx 2.87## s
• Common height: ##y \approx 11.4## m

Key Takeaways

• Model each projectile with its own kinematic equation.
• Account for delayed launches using shifted time expressions.
• Set vertical positions equal to find intersection times.
• Verify with both equations to confirm consistency.

Summary Table

Parameter	Stone A	Stone B
Initial Velocity (##v_0##)	18 m/s	32 m/s
Launch Time	##t = 0## s	##t = 2.5## s
Equation of Motion	##y_A(t) = 18t – 4.9t^2##	##y_B(t) = 32(t – 2.5) – 4.9(t – 2.5)^2##
Meeting Time	##t \approx 2.87## s
Common Height	##y \approx 11.4## m

Similar Problems — Quick Meeting Time Calculations

These quick solutions illustrate how to determine the time when two vertically launched objects reach the same height under gravity. Each case considers different launch velocities and possible launch delays.

Problem 1

Two balls thrown upwards simultaneously with initial velocities ##v_{01} = 15## m/s and ##v_{02} = 20## m/s. Meeting time: ≈ 0.51 s

Problem 2

Two rockets: Rocket A launched at ##t=0## with ##50## m/s, Rocket B launched 3 s later with ##60## m/s. Meeting time: ≈ 6.12 s

Problem 3

Stone A launched at ##t=0## with ##20## m/s, Stone B launched at ##t=1## s with ##25## m/s. Meeting time: ≈ 2.72 s

Problem 4

Two projectiles: velocities ##30## m/s and ##40## m/s, with a launch delay of 2 s. Meeting time: ≈ 4.39 s

Problem 5

Two objects thrown with velocities ##25## m/s and ##35## m/s, second launched 1.5 s later. Meeting time: ≈ 3.63 s

Summary Table

Problem	Launch Details	Meeting Time
1	v₁ = 15 m/s, v₂ = 20 m/s, same start	≈ 0.51 s
2	Rocket A: 50 m/s at t=0, Rocket B: 60 m/s at t=3 s	≈ 6.12 s
3	Stone A: 20 m/s at t=0, Stone B: 25 m/s at t=1 s	≈ 2.72 s
4	Projectile A: 30 m/s, Projectile B: 40 m/s, 2 s apart	≈ 4.39 s
5	Object A: 25 m/s, Object B: 35 m/s, 1.5 s apart	≈ 3.63 s

---

---